\(B=3x^2-2x+7\\ =3\left(x^2-\dfrac{2}{3}x+\dfrac{1}{9}\right)+\dfrac{20}{3}\\ =3\left(x-\dfrac{1}{3}\right)^2+\dfrac{20}{3}\\ Vì:\left(x-\dfrac{1}{3}\right)^2\ge0\Rightarrow3\left(x-\dfrac{1}{3}\right)^2\ge0\forall x\in R\\ Vậy:min_B=\dfrac{20}{3}khi.\left(x-\dfrac{1}{3}\right)=0\Leftrightarrow x=\dfrac{1}{3}\)